Problem 16 - College Algebra

Question

For the following exercises, find the \(x\) - or \(t\) -intercepts of the polynomial functions. $$ f(x)=x^{3}+2 x^{2}-9 x-18 $$

Step-by-Step Solution

Answer

The \(x\)-intercepts are \(x = 3\), \(x = -2\), and \(x = -3\).

Step 1: Set the function to zero

To find the intercepts, we need to set the function equal to zero and solve for \(x\). That is, we solve \(f(x) = 0\). Write down the equation:\[x^3 + 2x^2 - 9x - 18 = 0\]

Step 2: Use the Rational Root Theorem

To find possible rational roots, apply the Rational Root Theorem, which suggests that any rational solution is a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is \(-18\) and the leading coefficient is \(1\). Therefore, possible rational roots are \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\).

Step 3: Test possible roots

Substitute these possible roots into the polynomial to test which ones yield zero. Starting with \(x = -1\):\[f(-1) = (-1)^3 + 2(-1)^2 - 9(-1) - 18 = -1 + 2 + 9 - 18 = -8\]Since \(-8 eq 0\), \(x = -1\) is not a root. Continue checking the other possibilities. After testing, \(x = 3\) gives:\[f(3) = 3^3 + 2(3)^2 - 9(3) - 18 = 27 + 18 - 27 - 18 = 0\]So, \(x = 3\) is a root.

Step 4: Divide the polynomial by \(x - 3\)

Having found a root \(x = 3\), perform polynomial division of \(x^3 + 2x^2 - 9x - 18\) by \(x - 3\) to factor the polynomial. The quotient from division is \(x^2 + 5x + 6\).

Step 5: Factor the quadratic polynomial

Factor \(x^2 + 5x + 6\) into two binomials:\[x^2 + 5x + 6 = (x + 2)(x + 3)\].

Step 6: Solve for more roots

Set each factor of the polynomial to zero: \[x - 3 = 0\] gives \(x = 3\), \[x + 2 = 0\] gives \(x = -2\), and \[x + 3 = 0\] gives \(x = -3\).

Step 7: List all intercepts

The \(x\)-intercepts of the polynomial are the values of \(x\) that satisfy \(f(x) = 0\). From previous steps, these are \(x = 3\), \(x = -2\), and \(x = -3\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool in algebra that helps us find possible rational roots of a polynomial equation. It is particularly useful when dealing with polynomials that have integer coefficients. The theorem states that any rational root, or zero, of a polynomial with integer coefficients can be expressed as the fraction \( \frac{p}{q} \), where \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient.

For the polynomial \( f(x) = x^3 + 2x^2 - 9x - 18 \), the constant term is \(-18\) and the leading coefficient is \(1\).

• Factors of \( -18 \) include: \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \)
• Factors of \( 1 \) are simply \( \pm 1 \).

By dividing each factor of \(-18\) by \(\pm 1\), we get the same list of possible rational roots. These roots are potential solutions that we test to see which ones actually make the polynomial equal to zero. Utilizing this theorem narrows down the number of values we need to check, making the process more efficient and manageable.

Polynomial Division

Once a possible root of a polynomial is found, we can use polynomial division to simplify the polynomial and discover more roots. Polynomial division works similarly to long division with numbers. It allows us to divide the original polynomial by a binomial, typically in the form of \(x - r\), where \(r\) is a root.

After applying the Rational Root Theorem and identifying \(x = 3\) as a root for \(f(x) = x^3 + 2x^2 - 9x - 18\), we divide the polynomial by \(x - 3\).

The division simplifies the polynomial to a quadratic: \(x^2 + 5x + 6\). This process reveals the structure underneath the cubic equation, offering a clearer path to finding the remaining roots through further factoring or additional solutions.

By dividing the polynomial, not only have we confirmed \(x = 3\) is a root, but we've also reduced the degree of the polynomial, making it easier to handle.

Factoring Quadratics

Factoring quadratics is a key technique in algebra, especially after reducing a polynomial's degree. For any quadratic in the form \(ax^2 + bx + c\), the goal is to rewrite it as a product of two binomials. This helps to easily find the x-intercepts or roots by solving equations of the form \((x - p)(x - q) = 0\).

In our example, we've reduced the polynomial to \(x^2 + 5x + 6\) after the polynomial division. To factor this quadratic, we look for two numbers that multiply to \(6\) (the constant term) and add to \(5\) (the coefficient of \(x\)).

• The numbers \(2\) and \(3\) fit these criteria because \(2 \times 3 = 6\) and \(2 + 3 = 5\).

Thus, \(x^2 + 5x + 6\) can be factored into \((x + 2)(x + 3)\). By setting each factor to zero and solving for \(x\), we find two additional roots: \(x = -2\) and \(x = -3\).

Factoring quadratics can often be achieved through trial and error, or by more structured approaches like the quadratic formula, but understanding the number relationships simplifies the task.